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What is the equation of the line tangent to x^2+y^2=2 at the point (1,1)

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f32 | Student, Grade 9 | Honors

Posted June 22, 2013 at 2:19 PM via web

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What is the equation of the line tangent to x^2+y^2=2 at the point (1,1)

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justaguide | College Teacher | (Level 2) Distinguished Educator

Posted June 22, 2013 at 2:22 PM (Answer #1)

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The equation of the circle is x^2+y^2 = 2. The value of the derivative `dy/dx` at at point gives the slope of the tangent at that point.

`(d(x^2+ y^2))/dx = 0`

=> `2x + 2y*(dy/dx) = 0`

=> `dy/dx = -x/y`

At the point (1,1) the slope of the tangent is -1.

The equation of the tangent is `(y - 1)/(x - 1) = -1`

=> y - 1 = 1 - x

=> x + y - 2 = 0

The required equation of the tangent is x + y - 2 = 0

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